結果
問題 | No.1988 Divisor Tiling |
ユーザー |
|
提出日時 | 2022-06-24 21:47:09 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 3 ms / 2,000 ms |
コード長 | 4,603 bytes |
コンパイル時間 | 2,678 ms |
コンパイル使用メモリ | 148,924 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-11-08 17:40:47 |
合計ジャッジ時間 | 4,201 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 32 |
ソースコード
#include<vector>#include<set>#include<map>#include<queue>#include<string>#include<algorithm>#include<iostream>#include<bitset>#include<functional>#include<chrono>#include<numeric>#include<cstdio>#include<cstring>#include<cstdlib>#include<cassert>#include<cmath>#include<iomanip>#include<random>#include<ctime>#include<complex>#include<type_traits>#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace std;using namespace __gnu_pbds;typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;typedef long long LL;typedef double D;#define all(v) (v).begin(), (v).end()mt19937 gene(chrono::system_clock::now().time_since_epoch().count());typedef complex<double> Complex;#define fi first#define se second#define ins insert#define pb push_backinline char GET_CHAR(){const int maxn = 131072;static char buf[maxn],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxn,stdin),p1==p2)?EOF:*p1++;}inline int getInt() {int res(0);char c = getchar();while(c < '0') c = getchar();while(c >= '0') {res = res * 10 + (c - '0');c = getchar();}return res;}inline LL fastpo(LL x, LL n, LL mod) {LL res(1);while(n) {if(n & 1) {res = res * (LL)x % mod;}x = x * (LL) x % mod;n /= 2;}return res;}template<LL mod> struct Num {LL a;Num operator + (const Num & b) { return Num{(a + b.a) % mod}; }Num operator - (const Num & b) { return Num{(a - b.a + mod) % mod}; }Num operator * (const Num & b) { return Num{a * b.a % mod}; }Num operator / (const Num & b) { return Num{a * fastpo(b.a, mod - 2, mod) % mod}; }void operator += (const Num & b) {if((a += b.a) >= mod) a -= mod;}void operator -= (const Num & b) {if((a -= b.a) < 0) a += mod;}void operator *= (const Num & b) { a = a * b.a % mod; }void operator /= (const Num & b) { a = a * fastpo(b.a, mod - 2, mod) % mod; }void operator = (const Num & b) { a = b.a; }void operator = (const LL & b) { a = b; }};template<LL mod> ostream & operator << (ostream & os, const Num<mod> & a) {os << a.a;return os;}LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }inline string itoa(LL x, int width = 0) {string res;if(x == 0) res.push_back('0');while(x) {res.push_back('0' + x % 10);x /= 10;}while((int)res.size() < width) res.push_back('0');reverse(res.begin(), res.end());return res;}const int _B = 131072;char buf[_B];int _bl = 0;inline void flush() {fwrite(buf, 1, _bl, stdout);_bl = 0;}__inline void _putchar(char c) {if(_bl == _B) flush();buf[_bl++] = c;}inline void print(LL x, char c) {static char tmp[20];int l = 0;if(!x) tmp[l++] = '0';else {while(x) {tmp[l++] = x % 10 + '0';x /= 10;}}for(int i = l - 1; i >= 0; i--) _putchar(tmp[i]);_putchar(c);}typedef double C;struct P {C x, y;void scan() {double _x, _y;scanf("%lf%lf", &_x, &_y);x = _x; y = _y;}void print() {cout << '(' << x << ' ' << y << ')' << endl;}P operator + (const P & b) const { return P{x + b.x, y + b.y}; }P operator - (const P & b) const { return P{x - b.x, y - b.y}; }C operator * (const P & b) const { return x * b.y - y * b.x; }C operator % (const P & b) const { return x * b.x + y * b.y; }};P operator * (const C & x, const P & b) { return P{x * b.x, x * b.y}; }const int N = 300033;const int LOG = 20;const int mod = 1e9 + 7;const int inf = 1e9 + 7;int n, m;int dx[4] = {1, 0, -1, 0};int dy[4] = {0, 1, 0, -1};int rela[N];int getr(int x) {int p = x;while(rela[p] != p) p = rela[p];int p1 = p; p = x;while(rela[p] != p) {int p2 = rela[p];rela[p] = p1;p = p2;}return p1;}int main() {int n, m;scanf("%d%d", &n, &m);bool flag = false;//cout << ilogb(m) << endl;if(m != (1 << ilogb(m))) {flag = true;m = n / m;}vector<vector<int> > ans(m + 1, vector<int>(n / m + 1));for(int i = m; i >= 2; i /= 2) {for(int j = i; j > i / 2; j--) {for(int k = 1; k <= n / m; k++) {ans[j][k] = (i / 2) * (n / m);}}}int cl = 0;for(int i = n / m; ; ) {if(i % 2 != 0) break;else i /= 2;for(int k = 0; k < i; k++) {ans[1][++cl] = i;}}for(int i = 1; n % i == 0; i *= 2) {for(int k = 0; k < i; k++) ans[1][++cl] = i;}if(flag) {m = n / m;}for(int i = 1; i <= m; i++) {for(int j = 1; j <= n / m; j++) {printf("%d%c", flag ? ans[j][i] : ans[i][j], j == n / m ? '\n' : ' ');}}}